So, I'm trying to use induction to prove:

$\displaystyle \sum_{k=0}^n\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n},~~$ $\displaystyle ~~for~all~n \in \mathbb{N}$

Base case is true.

Assume:

$\displaystyle \left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}$

Test:

$\displaystyle \left(\frac{-1}{3}\right)^{k+1} = \frac{3}{4} + \frac{1}{4}(-1)^{(n+1)}3^{-(n+1)}$

$\displaystyle \left(\frac{-1}{3}\right)^k\left(\frac{-1}{3}\right) = \frac{3}{4} + \frac{1}{4}(-1)^{n}(-1)3^{-n}3^{-1}$

$\displaystyle \left(\frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}\right)\left(\frac{-1}{3}\right) = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

This next step is what seems to cause the problems. If the 3/4 on the LHS was not in brackets, everything would be fine and become = when I multiply the other term by -1/3; however, by multiplying the 3/4 by -1/3 it makes them not equal:

$\displaystyle -\frac{1}{4} - \frac{1}{12}(-1)^n3^{-n} = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

What am I doing wrong?